Question: A box contains $2$ red marbles, $3$ green marbles, and $3$ blue marbles. If we choose a marble, then another marble without putting the first one back in the box, what is the probability that the first marble will be green and the second will be blue?
The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened. In this case, event A is picking a green marble and leaving it out. Event B is picking a blue marble. Let's take the events one at at time. What is the probability that the first marble chosen will be green? There are $3$ green marbles, and $8$ total, so the probability we will pick a green marble is $\dfrac{3} {8}$. After we take out the first marble, we don't put it back in, so there are only $7$ marbles left. Since the first marble was green, there are still $3$ blue marbles left. So, the probability of picking a blue marble after taking out a green marble is $\dfrac{3} {7}$. Therefore, the probability of picking a green marble, then a blue marble is $\dfrac{3}{8} \cdot \dfrac{3}{7} = \dfrac{9}{56}$